Solve the equation $6^{z}(6^{z} - 1) = 12.$ A curve has the equation $y = x e^{3x}$. Find $\dfrac{dy}{dx}$. Find the value of $k$ for which $\dfrac{d^{2}y}{dx^{2}} = k e^{3x}(2 + 3x).$

1. Use substitution to form a quadratic equation

Let $u = 6^z$.
$u(u - 1) = 12$
$u^2 - u - 12 = 0$

2. Solve the quadratic equation

$(u - 4)(u + 3) = 0$
$u = 4$ or $u = -3$

3. Substitute back and reject invalid solutions

$6^z = 4$ or $6^z = -3$
Since $6^z > 0$ for all real $z$, $6^z = -3$ has no solution.

4. Solve for $z$ using logarithms

$6^z = 4$
$\ln(6^z) = \ln 4$
$z \ln 6 = \ln 4$
$z = \frac{\ln 4}{\ln 6} \approx 0.77370$
$z \approx 0.774$ (to 3 s.f.)

1. Apply the product rule to differentiate $x e^{3x}$

Let $u = x$ and $v = e^{3x}$.
Then $\frac{du}{dx} = 1$ and $\frac{dv}{dx} = 3e^{3x}$.
$\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$
$\frac{dy}{dx} = x(3e^{3x}) + e^{3x}(1)$

2. Factorise the expression to give the final answer

$\frac{dy}{dx} = e^{3x}(3x + 1)$

1. Recall or calculate the first derivative from part (b)(i)

From part (b)(i), using the product rule on $y = x e^{3x}$:
$\frac{dy}{dx} = (1)e^{3x} + x(3e^{3x}) = e^{3x}(1 + 3x)$

2. Differentiate the first derivative to find the second derivative

Using the product rule again on $\frac{dy}{dx} = e^{3x}(1 + 3x)$:
Let $u = e^{3x}$ and $v = 1 + 3x$.
Then $\frac{du}{dx} = 3e^{3x}$ and $\frac{dv}{dx} = 3$.
$\frac{d^{2}y}{dx^{2}} = 3e^{3x}(1 + 3x) + e^{3x}(3)$

3. Factorise the expression and compare to find $k$

$\frac{d^{2}y}{dx^{2}} = e^{3x}(3 + 9x + 3)$
$\frac{d^{2}y}{dx^{2}} = e^{3x}(6 + 9x)$
$\frac{d^{2}y}{dx^{2}} = 3e^{3x}(2 + 3x)$
Comparing this with $\frac{d^{2}y}{dx^{2}} = k e^{3x}(2 + 3x)$, we have:
$k = 3$